2011 SAC 1 - test

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12PHYSICS GENERATION Solutions

Victorian Certificate of Education 2017

NAME:_________________________TEACHER: ________________________

Physics

School Assessed Coursework: Test

Generation - Solutions

Reading Time: 10 minutes

Writing time: 60 minutes

QUESTION AND ANSWER BOOK

Section

Number of

questions

Number of questions

to be answered

Number of marks

A

4

4

41

· Students are permitted to bring into the examination room: pens, pencils, highlighters, erasers, sharpeners, rulers, up to two pages (one A4 sheet) of pre-written notes (typed or handwritten) and one scientific calculator.

· Students are NOT permitted to bring into the examination room: blank sheets of paper and/or white out liquid/tape.

Materials Supplied

· Question and Answer book of 10 pages. A formula sheet.

Instructions

· Write your student name and teacher name in the space provided above on this page.

· Unless otherwise indicated, the diagrams in this book are not drawn to scale.

· All written responses must be in English.

At the end of the examination

· Close your Question and Answer booklet and ensure your name and your teacher’s name is clearly printed on the front cover.

Students are NOT permitted to bring mobile phones and/or any other unauthorised electronic devices into the examination room.

Area of study – Generation

Instructions

Answer all questions in the spaces provided. Use black or blue pen.

Where an answer box has a unit printed in it, give your answer in that unit.

You should take the value of g to be 9.8 m s-2.

Where answer boxes have been provided, write your final answer in the box.

In questions where more than one mark is available, appropriate working should be shown.

Unless otherwise indicated, diagrams are not drawn to scale.

Question 1 (10 marks)

A single rectangular coil of side lengths 5.0 cm and 6.0 cm moves at a constant speed of 1.5 cm s-1. It moves from Position A outside of the magnetic field, to position B inside of the magnetic field, to position C again outside of the magnetic field. The magnetic field strength is 0.30 T.

C

A

B

15 cm

............................

5.0 cm

6.0 cm

1.5 cm s-1

a. Which is the magnitude of the magnetic flux when the coil is in position A?1 mark

_________________________________________________________________________

0 Wb

b. Show that magnetic flux in position B is 9 × 10-4 Wb.2 marks

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________________________________________________________________________________

________________________________________________________________________________

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c. On the axes provided, draw magnetic flux. Take t = 0 s to be the time when the coil first starts to enter the magnetic field.

2 marks

ΦB (×10-4 Wb)

d. Find the magnitude of the induced voltage across the terminals at t = 2 s.2 marks

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________________________________________________________________________________

________________________________________________________________________________

2.25 × 10-4 V

e. What is the induced voltage at t = 8 s. Explain your answer with reference to relevant physical laws3 marks

________________________________________________________________________________

· 0 V.

· Faraday’s Law says that induced voltage will occur when the magnetic flux threading the loop is changing.

· At t = 8 s, the magnetic flux through the loop is a constant

9 × 10-4 Wb and therefore will not induce a voltage across its terminals.

________________________________________________________________________________

________________________________________________________________________________

________________________________________________________________________________

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0 V

Question 2 (15 marks)

The diagram shows a generator with a square coil of side length 8 cm that rotates clockwise at 30 Hz. The magnetic field in which the coil rotates is 40 mT. The coil has 50 turns.

S

N

A

X

Y

a. What is the magnetic flux through the coil when the coil is in the position shown in the diagram? Include the unit with your answer. 1 mark

0 Wb

b. Calculate the magnitude of the average induced EMF between the slip rings as the coil rotates 90o clockwise from the position shown in the diagram.3 marks

________________________________________________________________________________

________________________________________________________________________________

________________________________________________________________________________

________________________________________________________________________________

________________________________________________________________________________

________________________________________________________________________________

________________________________________________________________________________

OR

1.54 V

c. In which direction does the induced current flow through the ammeter in the first quarter turn? Explain your answer.3 marks

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· As the coil rotates clockwise, the flux threading the loop increases to the left.

· Lenz’s law states the induced current will produce a flux threading the loop to the right in order to oppose the increasing flux to the left.

· Using the RH grip rule, the current must flow from X to Y

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________________________________________________________________________________

________________________________________________________________________________

________________________________________________________________________________

________________________________________________________________________________

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d. Draw the flux vs. time graph for the two full rotations of the coil from the position in the diagram. Include a scale on both the horizontal and vertical axis. 3 marks

Flux (×10-6 Wb)

e. Draw the induced EMF vs. time graph for the two full rotations of the coil from the position in the diagram. You do not need to include a scale on the induced EMF axis. 3 marks

f. Describe the changes to the induced EMF vs. time graph above if the coil was replaced with one that was the same size but had 75 turns and was rotated at 15 Hz.2 marks

________________________________________________________________________________

·

Amplitude will be of the original graph.

· The period will double.

________________________________________________________________________________

________________________________________________________________________________

________________________________________________________________________________

________________________________________________________________________________

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Question 3 (11 marks)

Farmer Joe uses running water from a river to turn a turbine that is connected to a generator. The generator is used to power his electric tools in the shed that is 3 km away and he has run an electrical cable that has a resistance of 0.003 Ω/m from the generator to the shed and back again. At the generator he measures a potential difference of 250 V but when he returns to test his power tools, the voltage drop at the shed is only 150 V and his tools do not run. Farmer Joe knows he requires at least 240 V in order for the power tools in his shed to run correctly.

3 km

R = 0.003 Ω m-1

V = 250 V

V = 150 V

a. Show that the current in the transmission lines is equal to 5.6 A.2 marks

________________________________________________________________________________

A

________________________________________________________________________________

________________________________________________________________________________

b. What is the power rating of the generator?2 marks

________________________________________________________________________________

________________________________________________________________________________

________________________________________________________________________________

________________________________________________________________________________

________________________________________________________________________________

1389 W

c. What power is dissipated during transmission?2 marks

OR

________________________________________________________________________________

________________________________________________________________________________

________________________________________________________________________________

________________________________________________________________________________

________________________________________________________________________________

556 W

Farmer Joe decides to use 2 transformers to reduce the power loss during transmission. He uses a 1:20 step-up transformer and a 20:1 step-down transformer.

d. Draw a diagram of Farmer Joe’s situation. Include the generator, transmission cables, the shed and the two transformers. Clearly indicate which transformer is step-up and step-down. 2 marks

Step Down

20:1

Step Up

1:20

3 km

R = 0.003 Ω m-1

V = 250 V

e. With the transformer arrangement, does Farmer Joe’s shed run the power tools as expected? Justify with calculations3 marks

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Voltage at the shed is 250-0.25 = 249.75 V

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Question 4 (5 marks)

The diagram below shows a transformer with 2400 turns in the primary coil and 300 turns in the secondary coil. The power supply runs at 110 V RMS.

~

V

a. Find the peak to peak voltage across the voltmeter.2 marks

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________________________________________________________________________________

________________________________________________________________________________

V

b. Explain, with reference to relevant physical laws, how a transformer works3 marks

________________________________________________________________________________

· The alternating current in the primary coil induces a changing flux in the soft iron core of the transformer.

· The changing flux is transmitted to the secondary coil via the soft iron core.

·

An EMF is induced across the secondary coil. Since the rate of change of flux is constant, the voltage across the secondary coil will depend on the number of turns. This is due to Faraday’s Law .

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END OF OUTCOME 2

END OF ANALYSIS TASK 2

2

L

PIR

=

2

5.55618

L

P

=´

20

T

S

V

V

D

D=

(

)

0.30.050.06

B

F=´´

5

20

S

V

D=

0.25 V

S

V

D=

5.556

20

T

I

=

0.278 A

T

I

=

T

VIR

D=

0.27818

T

V

D=´

5 V

T

V

D=

0.0009 Wb

B

F=

300110

2400

S

V

´

=

13.8 V

S

V

=

SS

PP

VN

VN

=

300

1102400

S

V

=

B

EMFN

t

DF

=-

D

4

9100

1

40

EMF

-

´-

=´

-

4

2.2510 V

EMF

-

=´

30 Hz

f

=

1

s

30

T

\=

1

turn

4

1

s

304

t

\=

´

1

turn

4

0.00833 s

t

=

(

)

2

3

40100.080

50

0.008330

EMF

-

-´´-

=-´

-

1.536 V

EMF

=

4

AV

EMFNfBA

=

(

)

(

)

2

3

4503040100.08

AV

EMF

-

=´´´´´

1.536 V

AV

EMF

=

7515

0.75

5030

´=

T

V

I

R

D

=

100

18

I

=

5.56

I

=

230000.003

T

R

=´´

18

T

R

=W

250150

V

D=-

100 V

V

D=

PVI

=

2505.556

P

=´

1389 W

P

=

B

BA

F=

2

L

V

P

R

D

=

2

100

18

L

P

=

556 W

L

P

=